1. Organic Spectroscopy
Mass Spectrometry

2. General
The mass spectrum is a plot of ion abundance versus m/e ratio (mass/charge ratio).
The most abundant ion formed in the ionization chamber gives to rise the tallest peak in the mass spectrum, called the base peak.
By using one of the many ionization methods, the simple removal of an electron from a molecule yields a positively charged radical cation, known as the molecular ion and symbolized as [M]+.
After formation of molecular Ion in the ionization chamber, excess energy causes further fragmentation of the molecular ion. Various positively charged masses (and/or positively charged radical cations) show up in the spectrum.
The mass of the molecular ion can be determined to an accuracy of ± of a mass unit, which yields a high resolution (hi-res) mass spectrum.
Ionization Methods
The most common method of ionization involves Electron impact (EI) in which molecule is bombarded with high-energy electrons. This is the strongest ionization method which usually causes further fragmentation. Since the molecular ion gets fragmented, the molecular ion usually produces a small peak with this method.

3. Mass Spectrum of Isobutyrophenone
105
(base peak)
C6H5CO+
molecular
weight = 148 77
C6H5+
molecular ion, M
(148)
M+1

4. Facts Concerning the Molecular Ion Peak
The peak must correspond to the ion of highest mass excluding the
usually much smaller isotopic peaks that occur at M+1, M+2, etc.
To be a molecular ion, the ion must contain an odd number of electrons. One electron is lost, forming a radical-cation.
To determine this, calculate the IHD. It must be a whole number.
Consider an ion at m/z = A possible molecular formula is C6H8O2.
The IHD = 3 (a whole number), so this could be the molecular ion.
However, an ion at 105 could correspond to a molecular formula of
C7H5O. The IHD is 5.5 (not a whole number), so this can’t be the
molecular ion.
The ion must be capable of producing smaller, fragment ions by loss
of neutral fragments of predictable structure.

5. The Lifetimes of Various Molecular Ions
Aromatics > conjugated alkenes > alicyclic compounds> organic sulfides >
unbranched hydrocarbons > mercaptans > ketones > amines > esters >
ethers > carboxylic acids > branched hydrocarbons > alcohols
Therefore, alcohols produce small or non-existent molecular ions because
their lifetimes are too short. They fragment before they can be detected.
CH3CH2CH2CH2OH
MW = 74
No M
visible

6. The Nitrogen Rule
The nitrogen rule states that an odd number of nitrogen atoms will form
a molecular ion with an odd mass number. An even number of nitrogen
atoms (or none at all) will produce a molecular ion with an even mass
number. This occurs because nitrogen has an odd-numbered valence.
Examples: C6H5CH2NH2 MW = 107
H2NCH2CH2NH2 MW = 60

7. Determining Possible Molecular Formulas from the Molecular Ion: Rule of 13
Rule of Thirteen: Based upon the assumption that CnHn and its mass of 13 is present in most organic compounds.
Divide the molecular ion by 13. This gives a value for n and any remainder (R) = additional H’s.
For a M+ = 106, n = 8(106/13) with a R of 2. A possible molecular formula for this ion is C8H8+2 = C8H10
For each CH there are heteroatom equivalents.

8. Rule of 13 Heteroatom Equivalents
Element
CH Equivalent
1H12 C
31P
C2H7
16O
CH4
32S
C2H8
14N
CH2
16O32S C4
16O14N
C2H6
35Cl
C2H11
19F
CH7
79Br
C6H7
28Si
C2H4
127I
C10H7

9. Candidate molecular formulas for M+. = 108
108/13 = 8, R = 4  C8H12
O = CH4; therefore, C8H12 – CH4 + O = C7H8O
or C8H12 – 2(CH4) + 2O = C6H4O2
or C8H12 – C6H7 + Br = C2H5Br
Calculate three candidate molecular formulas for C10H18.

10. When an odd amu M+. is seen, suspect one nitrogen or an odd
multiple. Candidate molecular formulas for a M+. = 121 are:
121/13 = 9, R = 4  C9H13; IHD = 3.5 so it can’t be M+.
O = CH4; N = CH2
or C9H13 – CH2 + N = C8H11N; IHD = 3, so it may be M+.
or C9H13 – 2(CH2) + 2N = C7H9N2; IHD = 4.5 so it can’t be M+.
or C9H13 – 3(CH2) + 3N = C6H7N3; IHD = 5, so it may be M+.
or C9H13 – (CH2) – (CH4) + N + O = C7H7NO IHD = 5,
so it may be M+.

11. Determining the Molecular Formula from the
Molecular Ion: Isotope Ratio Data
In this method, the relative intensities of the peaks due to the molecular ion
and related isotopic peaks are examined.
Advantage:
Does not require an expensive high-res MS instrument.
Disadvantages:
Isotopic peaks may be difficult to locate.
Useless when the molecular ion peak is very weak or does not appear.
3-pentanone

12. Relative Abundances of Common Elements and Their isotopes
Carbon
12C
100
13C
1.08
Hydrogen 1H 2H
0.016
Nitrogen
14N
15N
0.38
Oxygen
16O
17O
0.04
18O
0.20
Sulfur
32S
33S
0.78
34S
4.40
Chlorine
35Cl
 37Cl
32.5
Bromine
79Br
81Br
98.0

13. Example: 3-pentanone, C5H10O
%(M + 1) = 100 (M + 1)/M = 1.08 x # C atoms x # H atoms x # O atoms
= 1.08 x x x 1= 5.60
Actual spectrum: [1% (M +1)/17.4% (M)] x 100 = 5.75

14. Halogen M M + 2 M + 4 M + 6 Br 100 97.7 Br2 195.0 95.4 Br3 293.0 286.0
Relative Intensities of Isotope Peaks for Bromine and Chlorine
Halogen M
M + 2
M + 4
M + 6 Br
100
97.7
Br2
195.0
95.4
Br3
293.0
286.0
93.4 Cl
32.6
Cl2
65.3
10.6
Cl3
97.8
31.9
3.47
BrCl
130.0
Br2Cl
228.0
159.0
31.2
Cl2Br
163.0
74.4
10.4

16. 236
234
238

17. 112
114

18. 192
190
194

19. Determining the Molecular Formula from
the Molecular Ion: High Resolution MS (HRMS)
Using low resolution (LR) MS, you could not distinguish between the following
molecular formulas, each of which has a mass of 60:
C3H8O = (3 x 12) + (8 x 1) + 16 = 60
C2H8N2 = (2 x 12) + (8 x 1) + (2 x 14) = 60
C2H4O2 = (2 x 12) + (4 x 1) + (2 x 16) = 60
CH4N2O = 12 + (4 x 1) + (2 x 14) + 16 = 60
However, they can be distinguished using HRMS.

20. Precise Masses of Some Common Elements
Atomic Weight
Isotope
Mass
Hydrogen
1.0097 1H 2H
Carbon
12C
13C
Nitrogen
14N
15N
Oxygen
16O
17O
18O
Fluorine
19F
Silicon
28.086
28Si
29Si
30Si
Phosphorus
30.974
31P
Sulfur
32.064
32S
33S
34S
Chlorine
35.453
35Cl
37Cl
Bromine
79.909
79Br
81Br
Iodine
127I
Using precise masses:
C3H8O =
C2H8N2 =
C2H4O2 =
CH4N2O =

21. Fragmentation Patterns
Most common: one-bond cleavage to produce an odd-electron neutral fragment,
(radical, which is not detected) and an even-electron carbocation. Ease of frag-
mentation to form cations follows the scheme below:
CH3+ < RCH2+ < R2CH+ < R3C+ < CH2=CH-CH2+ < C6H5-CH2+
Difficult Easy
Radical (not detected)

22. Fragmentation Patterns (cont.)
Two-bond cleavage: The odd-electron molecular ion produces an odd-electron
fragment ion and an even-electron neutral fragment (not detected).
Not detected
McLafferty Rearrangement

23. Cleavage at branch points
MW = 86 57 71

24. -cleavage to hetero atoms
MW = 102
cleavage  to hetero atoms
CH3-CH=OH+ 43 87

25. -cleavage to aromatic ring
MW = 134 91
92 (from McLafferty
Rearrangement)

26. Cleavage  to carbonyl groups
MW = 86 43

27. McLafferty rearrangement carboxylic acids60

28. McLafferty rearrangement
esters
MW = 102 74

29. Hexane C6H14 MW = 86.18
Molecular ion peaks are present, possibly with low intensity. 
The fragmentation pattern contains clusters of peaks 14 mass
units apart (which represent loss of (CH2)nCH3).

30. 3-Pentanol C5H12O MW = 88.15
An alcohol's molecular ion is small or non-existent. 
Cleavage of the C-C bond next to the oxygen usually
occurs.  A loss of H2O may occur as in the spectrum below.

31. 3-Phenyl-2-propenal C9H8O MW = 132.16
Cleavage of bonds next to the carboxyl group
results in the loss of hydrogen (molecular ion less 1)
or the loss of CHO (molecular ion less 29).

32. 3-Methylbutyramide C5H11NO MW = 101.15
Primary amides show a base peak
due to the McLafferty rearrangement.

33. n-Butylamine C4H11N MW = 73.13 Molecular ion peak is an odd number.
Alpha-cleavage dominates aliphatic amines.

34. n-Methylbenzylamine C8H11N MW = 121.18
Another example is a secondary amine shown below. 
Again, the molecular ion peak is an odd number. 
The base peak is from the C-C cleavage adjacent to the C-N bond.

35. Molecular ion peaks are strong due to the stable structure.
Naphthalene C10H8 MW =
Molecular ion peaks are strong due to the stable structure.

36. 2-Butenoic acid C4H6O2 MW = 86.09
In short chain acids, peaks due to the loss of OH
(molecular ion less 17) and COOH (molecular ion less 45)
are prominent due to cleavage of bonds next to C=O.

37. Fragments appear due to bond cleavage next to C=O
(alkoxy group loss, -OR) and hydrogen rearrangements.                                                 
Ethyl acetate C4H8O2 MW = 88.11

38. Ethyl methyl ether C3H8O MW = 60.10
Fragmentation tends to occur alpha to the
oxygen atom (C-C bond next to the oxygen).

39. 1-Bromopropane C3H7Br MW = 123.00
The presence of chlorine or bromine atoms is
usually recognizable from isotopic peaks.

40. 4-Heptanone C7H14O MW =
Major fragmentation peaks result from cleavage
of the C-C bonds adjacent to the carbonyl.

41. Which structure supports the following mass spectrum?

42. Which structure supports the following mass spectrum?

43. Which structure supports the following mass spectrum?

44. An unknown compound has the mass spectrum shown below. The IR spectrum
shows peaks in the and the cm-1 ranges and a strong
absorption at 1688 cm-1. Suggest a structure consistent with this data.

45. An unknown compound has the mass spectrum shown below
An unknown compound has the mass spectrum shown below. The IR spectrum shows peaks in the cm-1 range and a strong absorption at 1718 cm-1. Suggest a structure consistent with this data.